Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>

using namespace std;

#define rep(i,n) for(int i=0;i<(int)n;i++)
#define int long long

vector<int> G[100000];

vector<int> bfs(int n,int s){
  vector<int> dist(n,1e9),visited(n);
  dist[s]=0;
  visited[s]=1;
  queue<int> que;
  que.push(s);
  while(!que.empty()){
    int i = que.front();
    que.pop();
    rep(j,G[i].size()){
      int to = G[i][j];
      if (visited[to]!=1) {
        visited[to]=1;
        dist[to] = dist[i]+1;
        que.push(to);
      }
    }
  }
  return dist;
}

signed main(){

  int n,m,s,t;
  cin >> n >> m >> s >> t;
  s--;
  t--;
  rep(i,m){
    int x,y;
    cin >> x >> y;
    x--;
    y--;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  vector<int> dist = bfs(n,s);
  vector<int> dist_r = bfs(n,t);
  int min_dist = dist[t];

  vector<int> dd(min_dist,0),dd_r(min_dist,0);
  rep(i,n){
    if (dist[i]<=min_dist-2)dd[dist[i]]++;
    if (dist_r[i]<=min_dist-2)dd_r[dist_r[i]]++;
  }
  long long sum = 0;
  for(int i=0;i<min_dist-1;i++){
    sum += dd[i]*dd_r[min_dist-2-i];
  }
  cout << sum << endl;

  return 0;
}